## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $L = 30~m$ (b) The new current is $1.0~A$
(a) We can find the resistance: $R = \frac{V}{I}$ $R = \frac{1.5~V}{0.50~A}$ $R = 3.0~\Omega$ We can find the required length: $R = \frac{\rho~L}{A}$ $L = \frac{R~A}{\rho}$ $L = \frac{R~\pi~r^2}{\rho}$ $L = \frac{(3.0~\Omega)~(\pi)~(0.30\times 10^{-3}~m)^2}{2.8\times 10^{-8}~\Omega~m}$ $L = 30~m$ (b) If the wire is half this length, then the resistance $R$ is also half the original value. Then, according to $V = IR$, the current must double since the voltage remains the same. The new current is $1.0~A$